I am sure there is a major flaw in my proof, but I cannot find it. I would be very happy if someone could point it out for me. Note that I am not looking for any hints or solutions. Below, $LR$ and $UR$ mean lower Riemann integral and upper Riemann integral.
$\textbf{Problem.}$ Let $\{f_{n}\}$ be a sequence of bounded functions that converge uniformly to $f$ on $[a,b]$. If each $f_n$ is Riemann integrable over $[a,b]$, show that $f$ is also Riemann integrable over $[a,b]$.
$\textbf{Proof.}$ We always have $LR\int_{a}^{b}f\leq UR\int_{a}^{b}f$. Let $\epsilon>0$. Then there exists an $N$ such that $f_{n}-\epsilon<f<f_{n}+\epsilon$ for all $n\geq N$ and $x\in [a,b]$. So if $n\geq N$, then
$LR\int_{a}^{b}f=$sup$\{\int_{a}^{b}\phi$ | $\phi$ is a step function and $\phi\leq f\}\geq$ sup$\{\int_{a}^{b}\phi$ | $\phi$ is a step function and $\phi\leq f_{n}-\epsilon\}=\int_{a}^{b}(f_{n}-\epsilon)=\int_{a}^{b}f_{n}-\epsilon(b-a)$.
Since $\epsilon$ is arbitrary we have $LR\int_{a}^{b}f\geq\int_{a}^{b}f_{n}$. Similarly, using $f\leq f_{n}+\epsilon$, we get $UR\int_{a}^{b}f\leq\int_{a}^{b}f_{n}$. Hence $LR\int_{a}^{b}f\geq UR\int_{a}^{b}f$, which completes the proof.
$\epsilon$ starts out arbitrary, but at this point of the proof, you have fixed $\epsilon$, and chosen an $N$ and an $n$ based on it. You can let $\epsilon$ go to zero, but then you have a different $f_n$. So you can't say $LR\int_{a}^{b}f\geq\int_{a}^{b}f_{n}$. You have to say $\lim_{n \rightarrow \infty} (LR\int_{a}^{b}f-\int_{a}^{b}f_{n})\geq0$.